Patrick Stevens

This post is for posterity, made shortly after Dr Paul Russell lectured Analysis II in Part IB of the Maths Tripos at Cambridge. In particular, he demonstrated a way of doing certain basic questions. It may be useful to people who are only just starting the study of analysis and/or who are doing example sheets in it.

The first example sheet of an Analysis course will usually be full of questions designed to get you up and running with the basic definitions. For instance, one question from the first example sheet of Analysis II this year is as follows:

Show that if is a sequence of uniformly continuous real functions on , and if uniformly, then is uniformly continuous.

This is one of those questions which only exists to make sure that you know what “uniformly continuous” and “converges uniformly” mean.

How do we solve this question? The key with a definitions-question is to avoid employing the brain wherever possible. So the first step is to define and , and to write down everything we know about them:

Let be a sequence of uniformly continuous real functions on , and a real function on , such that uniformly.

Since each is uniformly continuous, we have that for all , we have for every there is a such that for all with , it is true that .

Since uniformly, we have that for all , there exists such that for all , for every we have .

Now, what do we want to prove?

[Don’t write this down yet - this line goes at the end of the proof!] Therefore, for every there is a such that for all with , . Hence is uniformly continuous.

So what can we get from what we know? Everything we know is about “for all ”. So we fix an arbitrary . If we can prove something that is true for this , with no further assumptions, then we are done for all .

Fix arbitrary greater than .

Now what do we know?

Since each is uniformly continuous, we have that for all , there is a such that for all with , it is true that .

Since uniformly, we have that there exists such that for all , for every we have .

.

Aha! Now we have a definite something existing (namely, the in the second condition). Let’s fix it into existence.

Let be such that for all , for every we have .

What do we know?

Since each is uniformly continuous, we have that for all , there is a such that for all with , it is true that .

Since uniformly, we have that for all , for every we have .

, and is an integer.

Now, we have two “for all”s competing with each other. The more specific is the second one, so we’ll fix that into existence.

Fix arbitrary greater than or equal to .

What do we know?

Since each is uniformly continuous, we have that for all , there is a such that for all with , it is true that .

Since uniformly, we have that for every , .

, and is an integer, and .

Now we have a choice of “for all”s again, but this time they aren’t “talking about the same thing” (last time, both were integers referring to which we were talking about; this time, one is an integer and one is an arbitrary real). However, now we have which we can talk about; let’s wring more information out of it, by using the “uniformly continuous” bit.

Since each is uniformly continuous, there is a such that for all with , it is true that .

Since uniformly, we have that for every , .

, and is an integer, and .

Aha - another “there exists” condition (on ). Let’s fix it.

Fix such that for all with , it is true that .

What do we know?

Since each is uniformly continuous, for all with , it is true that .

Since uniformly, we have that for every , .

, and is an integer, and , and .

Two more “for all” conditions. Let’s fix them into existence:

Let be an arbitrary real, and let be such that .

What do we know?

Since each is uniformly continuous, .

Since uniformly, we have that .

, and is an integer, and , and , and is real, and .

Now the conditions are really small things. It’s kind of unclear how to proceed from here, so let’s look at what we wanted to prove again:

For every there is a such that for all with , .

Applying what we know, this becomes:

[to be proved] For all with , .

Aha! We have already got something to do with (namely that ), and we have something to do with (namely that ). Hence , and the triangle inequality gives us that . Eek - we need to turn that into an . We have no way of doing that, so we must have missed out some information somewhere. Backtracking, the nearest-to-the-end bit of missed out information was when we fixed . We threw away information in “for every , ” when we fixed - it applies to too. So we’ll add a new statement to the “what do we know?” list:

.

, and is an integer, and , and , and is real, and .

And now it just drops out of the triangle inequality that .

Now, was arbitrary, was dictated by the conditions, was arbitrary, was dictated by the conditions, was arbitrary, was arbitrary subject to .

Hence we have proved that for every there exists such that for all there is a such that for all , for all with , .

We can clean this statement up. Notice that neither nor was involved in the final expression, so we can simply get rid of them to obtain:

For every there is a such that for all , for all with , .

From this, it is easy to obtain the required result. We want to turn into - but that’s fine, because the expression holds for every , so in particular if we fix then it holds for . We’ll just use the from that instead. This gives us that is uniformly continuous, as required, and without actually engaging the brain except to carry out the algorithm of “write down what we know; if there exists something, fix it, and repeat; if for all something, then fix an arbitrary one, and repeat; if we’re stuck, go back through, looking to see if we missed out any information during a fixing-arbitrary-for-all phase” and to carry out the algorithm of “when the information we have is simple enough, compare terms from what we know with the expression that we want to show; use the triangle inequality to get them in there”.